\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [63]

Optimal result

Integrand size = 33, antiderivative size = 192 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 A+2 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (76 A+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/2*(13*A+2*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(76*A+11*C)*tan(d*x+c)/a^3/d+1/2*(13*A+2*C)*sec(d*x+c)*tan(d*x+c
)/a^3/d-1/5*(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(11*A+C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(
d*x+c))^2-1/15*(76*A+11*C)*sec(d*x+c)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3121, 3057, 2827, 3853, 3855, 3852, 8} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 A+2 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (76 A+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {(76 A+11 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(11 A+C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]

[Out]

((13*A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(76*A + 11*C)*Tan[c + d*x])/(15*a^3*d) + ((13*A + 2*C)*Sec
[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A + C)*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((11*A
+ C)*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((76*A + 11*C)*Sec[c + d*x]*Tan[c + d*x])/(1
5*d*(a^3 + a^3*Cos[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3121

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x
])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (7 A+2 C)-a (4 A-C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (43 A+8 C)-3 a^2 (11 A+C) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (15 a^3 (13 A+2 C)-2 a^3 (76 A+11 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{15 a^6} \\ & = -\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(13 A+2 C) \int \sec ^3(c+d x) \, dx}{a^3}-\frac {(2 (76 A+11 C)) \int \sec ^2(c+d x) \, dx}{15 a^3} \\ & = \frac {(13 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(13 A+2 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac {(2 (76 A+11 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d} \\ & = \frac {(13 A+2 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (76 A+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(597\) vs. \(2(192)=384\).

Time = 4.17 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.11 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {1920 (13 A+2 C) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (-5 (247 A+98 C) \sin \left (\frac {d x}{2}\right )+5 (761 A+106 C) \sin \left (\frac {3 d x}{2}\right )-4329 A \sin \left (c-\frac {d x}{2}\right )-654 C \sin \left (c-\frac {d x}{2}\right )+1989 A \sin \left (c+\frac {d x}{2}\right )+654 C \sin \left (c+\frac {d x}{2}\right )-3575 A \sin \left (2 c+\frac {d x}{2}\right )-490 C \sin \left (2 c+\frac {d x}{2}\right )-475 A \sin \left (c+\frac {3 d x}{2}\right )-350 C \sin \left (c+\frac {3 d x}{2}\right )+2005 A \sin \left (2 c+\frac {3 d x}{2}\right )+530 C \sin \left (2 c+\frac {3 d x}{2}\right )-2275 A \sin \left (3 c+\frac {3 d x}{2}\right )-350 C \sin \left (3 c+\frac {3 d x}{2}\right )+2673 A \sin \left (c+\frac {5 d x}{2}\right )+378 C \sin \left (c+\frac {5 d x}{2}\right )+105 A \sin \left (2 c+\frac {5 d x}{2}\right )-150 C \sin \left (2 c+\frac {5 d x}{2}\right )+1593 A \sin \left (3 c+\frac {5 d x}{2}\right )+378 C \sin \left (3 c+\frac {5 d x}{2}\right )-975 A \sin \left (4 c+\frac {5 d x}{2}\right )-150 C \sin \left (4 c+\frac {5 d x}{2}\right )+1325 A \sin \left (2 c+\frac {7 d x}{2}\right )+190 C \sin \left (2 c+\frac {7 d x}{2}\right )+255 A \sin \left (3 c+\frac {7 d x}{2}\right )-30 C \sin \left (3 c+\frac {7 d x}{2}\right )+875 A \sin \left (4 c+\frac {7 d x}{2}\right )+190 C \sin \left (4 c+\frac {7 d x}{2}\right )-195 A \sin \left (5 c+\frac {7 d x}{2}\right )-30 C \sin \left (5 c+\frac {7 d x}{2}\right )+304 A \sin \left (3 c+\frac {9 d x}{2}\right )+44 C \sin \left (3 c+\frac {9 d x}{2}\right )+90 A \sin \left (4 c+\frac {9 d x}{2}\right )+214 A \sin \left (5 c+\frac {9 d x}{2}\right )+44 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \]

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/480*(1920*(13*A + 2*C)*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-5*(247*A + 98*C)*Sin[(d*x)/2] + 5*(76
1*A + 106*C)*Sin[(3*d*x)/2] - 4329*A*Sin[c - (d*x)/2] - 654*C*Sin[c - (d*x)/2] + 1989*A*Sin[c + (d*x)/2] + 654
*C*Sin[c + (d*x)/2] - 3575*A*Sin[2*c + (d*x)/2] - 490*C*Sin[2*c + (d*x)/2] - 475*A*Sin[c + (3*d*x)/2] - 350*C*
Sin[c + (3*d*x)/2] + 2005*A*Sin[2*c + (3*d*x)/2] + 530*C*Sin[2*c + (3*d*x)/2] - 2275*A*Sin[3*c + (3*d*x)/2] -
350*C*Sin[3*c + (3*d*x)/2] + 2673*A*Sin[c + (5*d*x)/2] + 378*C*Sin[c + (5*d*x)/2] + 105*A*Sin[2*c + (5*d*x)/2]
- 150*C*Sin[2*c + (5*d*x)/2] + 1593*A*Sin[3*c + (5*d*x)/2] + 378*C*Sin[3*c + (5*d*x)/2] - 975*A*Sin[4*c + (5*
d*x)/2] - 150*C*Sin[4*c + (5*d*x)/2] + 1325*A*Sin[2*c + (7*d*x)/2] + 190*C*Sin[2*c + (7*d*x)/2] + 255*A*Sin[3*
c + (7*d*x)/2] - 30*C*Sin[3*c + (7*d*x)/2] + 875*A*Sin[4*c + (7*d*x)/2] + 190*C*Sin[4*c + (7*d*x)/2] - 195*A*S
in[5*c + (7*d*x)/2] - 30*C*Sin[5*c + (7*d*x)/2] + 304*A*Sin[3*c + (9*d*x)/2] + 44*C*Sin[3*c + (9*d*x)/2] + 90*
A*Sin[4*c + (9*d*x)/2] + 214*A*Sin[5*c + (9*d*x)/2] + 44*C*Sin[5*c + (9*d*x)/2]))/(a^3*d*(1 + Cos[c + d*x])^3)

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.91

method	result	size
parallelrisch	\(\frac {-1560 \left (A +\frac {2 C}{13}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+1560 \left (A +\frac {2 C}{13}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-152 \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {783 A}{76}+\frac {27 C}{19}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {717 A}{152}+\frac {51 C}{76}\right ) \cos \left (3 d x +3 c \right )+\left (A +\frac {11 C}{76}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {2331 A}{152}+\frac {153 C}{76}\right ) \cos \left (d x +c \right )+\frac {677 A}{76}+\frac {97 C}{76}\right )}{240 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(175\)
derivativedivides	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (26 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-26 A -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{4 d \,a^{3}}\)	\(194\)
default	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (26 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-26 A -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{4 d \,a^{3}}\)	\(194\)
norman	\(\frac {-\frac {\left (A +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (2 A +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (11 A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (25 A +2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (51 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {3 \left (51 A +11 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {3 \left (163 A +23 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a^{2}}-\frac {\left (13 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (13 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\)	\(263\)
risch	\(-\frac {i \left (195 A \,{\mathrm e}^{8 i \left (d x +c \right )}+30 C \,{\mathrm e}^{8 i \left (d x +c \right )}+975 A \,{\mathrm e}^{7 i \left (d x +c \right )}+150 C \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 A \,{\mathrm e}^{6 i \left (d x +c \right )}+350 C \,{\mathrm e}^{6 i \left (d x +c \right )}+3575 A \,{\mathrm e}^{5 i \left (d x +c \right )}+490 C \,{\mathrm e}^{5 i \left (d x +c \right )}+4329 A \,{\mathrm e}^{4 i \left (d x +c \right )}+654 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3805 A \,{\mathrm e}^{3 i \left (d x +c \right )}+530 C \,{\mathrm e}^{3 i \left (d x +c \right )}+2673 A \,{\mathrm e}^{2 i \left (d x +c \right )}+378 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1325 A \,{\mathrm e}^{i \left (d x +c \right )}+190 C \,{\mathrm e}^{i \left (d x +c \right )}+304 A +44 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}+\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}\)	\(323\)

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/240*(-1560*(A+2/13*C)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+1560*(A+2/13*C)*(1+cos(2*d*x+2*c))*ln(tan(
1/2*d*x+1/2*c)+1)-152*sec(1/2*d*x+1/2*c)^4*tan(1/2*d*x+1/2*c)*((783/76*A+27/19*C)*cos(2*d*x+2*c)+(717/152*A+51
/76*C)*cos(3*d*x+3*c)+(A+11/76*C)*cos(4*d*x+4*c)+(2331/152*A+153/76*C)*cos(d*x+c)+677/76*A+97/76*C))/d/a^3/(1+
cos(2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.51 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (76 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, A + 34 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, A + 64 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, A \cos \left (d x + c\right ) - 15 \, A\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((13*A + 2*C)*cos(d*x + c)^5 + 3*(13*A + 2*C)*cos(d*x + c)^4 + 3*(13*A + 2*C)*cos(d*x + c)^3 + (13*A
+ 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((13*A + 2*C)*cos(d*x + c)^5 + 3*(13*A + 2*C)*cos(d*x + c)^4
+ 3*(13*A + 2*C)*cos(d*x + c)^3 + (13*A + 2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(76*A + 11*C)*co
s(d*x + c)^4 + 3*(239*A + 34*C)*cos(d*x + c)^3 + (479*A + 64*C)*cos(d*x + c)^2 + 45*A*cos(d*x + c) - 15*A)*sin
(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.72 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {A {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(A*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + C*((105*sin(d*x + c)/(cos(d*x
+ c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(
d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (13 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(13*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(7*A*tan(1/2*d*x + 1/2*c)^3 - 5*A*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3
*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 20*C*a^1
2*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.02 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {13\,A}{2}+C\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^3}+\frac {3\,\left (5\,A+C\right )}{4\,a^3}+\frac {10\,A-2\,C}{4\,a^3}\right )}{d}-\frac {5\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^3}+\frac {5\,A+C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^3),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*((13*A)/2 + C))/(a^3*d) - (tan(c/2 + (d*x)/2)*((3*(A + C))/(2*a^3) + (3*(5*A + C)
)/(4*a^3) + (10*A - 2*C)/(4*a^3)))/d - (5*A*tan(c/2 + (d*x)/2) - 7*A*tan(c/2 + (d*x)/2)^3)/(d*(a^3*tan(c/2 + (
d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^3) + (5*A + C)/(12*a^3)))
/d - (tan(c/2 + (d*x)/2)^5*(A + C))/(20*a^3*d)